Markus Hanke

In some sense, yes. But it isn’t really a physical change that one would notice - spacetime remains smooth, regular, and locally Minkowskian everywhere (outside the singularity). What changes is mostly the physical meaning of the coordinates we use, relative to the exterior region. They now become spatial in nature, along the radial direction. Future means going “down” radially, past means going “up”. So the physical meaning of the r and t coordinates trade places. But again, it’s not something you would notice; spacetime looks just the same below as above the horizon. The only difference is its causal structure - below the horizon, all physically possible world lines (whether geodesics or not) terminate at the singularity. Locally, everything looks perfectly normal there, at least up to the point where tidal forces become noticeable. No, nothing special happens at the horizon at all - spacetime is perfectly regular there, and if you were to fall through it, you wouldn’t locally notice anything out of the ordinary. It’s really just a mathematical concept, not a physical entity. He can, because he’s in free fall. Visualise it like this (though it’s not really correct) - a photon emitted radially outwards very close to the horizon has a very slow radial (!) velocity wrt to the event horizon. On the other hand though, Pinocchio falls through the horizon at nearly the speed of light (wrt some outside reference), and thus meets the photon on the way. So it’s not like the photon necessarily propagates to his eyes, but rather that his eyes fall right to where the photon is. It’s kind of like jumping upwards in an elevator - you can put relative motion between yourself and the elevator floor, but both you and the elevator continue to move down regardless (maybe a stupid example, but you get my drift hopefully). Note that the situation is different if you’re not in free-fall. If Pinocchio, after the tip of his nose crosses the horizon, somehow fires magical thrusters that arrest his fall before he reaches the EH, then his nose will visually disappear for him (and get ripped away).

The diagram is necessarily correct (it shows a valid solution to the EFE), and I think so is Genady’s interpretation of it. The thing is that, at the event horizon, something very unintuitive happens - the physical meaning of the coordinates we use is no longer the way we are accustomed to. Imagine an astronaut in free fall, just as he crosses the horizon - let’s for simplicity’s sake say his feet emit light. Once his feet have crossed the horizon, and always assuming free fall, this light signal is now no longer “below” the eyes, but in their future. Light below the horizon is perfectly free to propagate in all spatial directions, yet it can still never leave the BH, because the singularity is in the future, and the horizon is in the past. It is no longer a question of up, down, above or below, once you’re past the horizon. Thus, for the astronaut, the light leaves his feet, and his eyes will necessarily “meet” it, because he’s in free fall. Both age towards the singularity, their relative velocity remains c (so everything is locally Minkowskian), yet their geodesics must intersect, just as the diagram shows. Thus he sees his feet like normal, perhaps slightly redshifted and dimmed. He will otherwise never notice anything special at the horizon. And he can’t, because locally everything must look Minkowskian. This is probably the biggest mistake people make when trying to visualise black holes - they think that, past the horizon, the singularity is “down”; but it’s not, it’s in the future. Likewise, the horizon isn’t “up”, but in the past. This is extremely important to understand, or else there’ll be all sorts of misunderstandings. It’s the other way around, see also above - light past the horizon remains past the horizon, but the eyes which see that light are falling inwards, and can intersect that light in the future.

It seems to be you who’s disagreeing with standard textbook stuff, such as spin, not me.

I’m a bit confused here - over on the other thread on cosmology you seemed to be implying that the theory of relativity is not a good model; yet here you talk about spin, which is a relativistic phenomenon? In the case of spin, this principle says that you cannot measure more than one component of the spin vector simultaneously with arbitrary precision. You can, however, measure one component plus the overall magnitude of the spin vector simultaneously without problems.

The question is one of scale, not balance. If you use GR to model any gravitational mechanics on a scale of the solar system, or some few multiples of it, you get the correct results to very high levels of accuracy - so the equation isn’t “flawed” in any meaningful sense. Remember that we have tested it very extensively locally here in the solar system. Rather, what happens is that on large scales, systems behave as if they contain much more matter than is visible in the electromagnetic spectrum. Fundamentally, this can mean one of three things: 1. There’s extra stuff there which we can’t see (dark matter) 2. There’s nothing extra there, but the laws of gravity have to be modified on larger scales; GR remains perfectly valid on solar scales 3. There may be some kind of other interaction happening, over and above gravity, which we don’t know about. So whatever happens, GR will remain a valid and good model; at most, its domain of applicability might become more limited.

Oh yes, there is a fundamental connection between these, given by Noether’s theorem - translation invariance in time corresponds to a conserved quantity, which is precisely the energy-momentum tensor. Without time, there would be no meaningful notion of energy-momentum. This is wrong - it’s called mass-energy equivalence, because there’s no distinction between them; mass is just a specific form of energy, they are equivalent to one another. It’s the other way around - energy-momentum arises (as a meaningful concept) from the continuous symmetries of this spacetime, in this case time-translation invariance and rotational invariance, via Noether’s theorem.

I call them by their usual names, Dark Matter and Dark Energy. Had you clearly stated that this is what you were referring to, your posts would have been easier to decipher. Nonetheless, the answer is the same as with the quantum gravity issue - right now we’re not sure about the precise nature of these entities, but it’s being worked on. Such things take time and effort to understand. Perhaps also the answer might be a modification of the laws of gravity (also being worked on); though, considering latest results, the air seems to be getting a bit thin for that option. Like I said, science is an ongoing process.

But we already have this? It’s called the Navier-Stokes equations, and they work pretty well.

What do you mean by this, exactly? The universe is just there - all we do in physics is to find models that provide the best possible descriptions of aspects of it. Most of it “unites” just fine, it’s only gravity that is a problem right now. But we’re working on this - physics, like any other science, is a process.

This is absolutely untrue, on all levels.

You are right in that it constrains the set of all possible global topologies, but what I attempted to point out is that it doesn’t uniquely determine it, at least not in 4D. More precisely, globally different topologies can be compatible with the same local geometry, so local measurements of curvature alone don’t necessarily give this information. The reverse is also true - manifolds can have the same global topology, yet different geometries. So this issue is subtle. The interesting exception is in 2D - here, the Ricci scalar is also the Euler characteristic, so gravity is entirely topological. A bit more info here: https://en.m.wikipedia.org/wiki/Shape_of_the_universe

This is true of course. It should be noted though that if both GR and QFT are valid theories, at least near the event horizon, then the existence of Hawking radiation is inevitable. If it turns out to not exist, then one or both of these models don’t apply in that region.

They didn’t know whether light has mass or not, it was an unanswered question at the time. But since Newtonian gravity can only handle test particles with mass, they had to assume that it did in order to derive any kind of prediction at all. The deflection angle doesn’t depend on the exact value of the mass - it just can’t be zero in Newton’s theory. You have to assume this to derive a deflection angle from Newton’s theory, which was the only theory available back then - it can’t handle massless test particles. You are absolutely right of course - this is one of the areas where Newtonian theory fails, and GR is needed. One must remember though that this wasn’t well understood prior to Einstein. Actually, it doesn’t - they are separate concepts. GR determines only local geometry, but not global topology. For example, the maximally extended Schwarzschild metric could describe both two separate, singly-connected regions of spacetime, or a single multiply-connected spacetime. Geometry is the same in both cases, but the global topology isn’t.

Newtonian gravity has nothing to say about massless particles, so strictly speaking it makes no prediction here. However, if one assumes that photons have a very small but finite mass, then one can use Newtonian gravity to work out how they are deflected around massive bodies. Turns out that deflection angle doesn’t depend on the exact mass of the photon, so long as it is much smaller than that of the central body. The result you get is off by a factor of 2 compared to actual observations - to get the correct angle, one must use GR.

Yes, this is possible; this is in fact one of the standard ways to (in principle) build GW detectors: https://arxiv.org/abs/1501.00996 You need an extended array of clocks for this, since what you are measuring is the dilation between clocks at different positions within a passing GW wavefront. Let’s just say it’s equally hard You’d need an extended array of very precise and perfectly synchronised atomic clocks. This is doable at least in principle with current technology.

No it’s not, because measurements of space and time are inherently observer-dependent concepts - there is no absolute frame at all, so there cannot be a contradiction. What all observers agree on is the spacetime interval. This is a rather silly statement, since whatever device you have used to create this post is based on a relativistic theory - the Standard Model, especially the part of it dealing with electromagnetism. Obviously, your computer isn’t an optical illusion, and using relativity to construct it has resulted in quite a useful machine. Have you ever used a microwave? An old-style CRT Monitor? Had an MRI scan? Seen a thermometer field with mercury? Used the GPS on your phone? Used electricity generated in a nuclear power station? Etc. All of these are things that inherently rely on relativistic effects to work.

By “geometry” I mean how exactly the region of spacetime in question is curved. There’s mainly two considerations that are of relevance in this context here - does spacetime become approximately flat if we go far enough away (asymptotic flatness), or are there distant sources of gravity that need to be accounted for? And secondly - what symmetries does this spacetime have? Can we translate each point within this region a short distance along space and/or time, without affecting any of the physics? For example, each point in ordinary Schwarzschild spacetime can be shifted along a vector pointing - say - 1 second into the future, without changing anything about the physics of the system - it thus is said to admit a time-like Killing field, which is to say it is a stationary spacetime. Killing fields are one way to speak about symmetries in spacetime. And since continuous symmetries are by Noether’s theorem associated with conserved quantities, this will have an impact on how we define concepts such as energy-momentum across an extended region. Different definitions are available for different types of spacetime exhibiting different symmetries; and if you’re in a spacetime that’s complicated enough so that it has very few or no symmetries, then it may not even be possible to define its global mass in a meaningful way at all. Does this make more sense?

No. What it means is that the answer to the question of “how much mass/energy is in a region of spacetime” depends on what kind of geometry that region has. Depending on considerations such as symmetries (Killing fields), asymptotic flatness etc a certain definition may apply, while other definitions may not work. So one has to be very careful which one is to be used. Note also that being in relative motion wrt to a gravitational source does not change the geometry of spacetime, it only changes the coordinate description of it.

It has similar challenges Try for example the word “Eichhörnchennachwuchs“ (=offspring of squirrels), where you have three entirely different sounds for the combination “ch”, all in a single word. You get the idea. Fair play! German is quite tough grammar-wise, whereas Chinese is simple in terms of structure, but requires lots of time and effort in memorising characters. I lived and worked in China for a year when I was younger, and immensely enjoyed learning it…but never used it afterwards, and now, 25 years later, I’ve forgotten pretty much everything 😕 You loose what you don’t use. Couldn’t agree more! I love teaching myself languages - I find it a very useful way to keep the old brain in shape, and it also helps to break up old deeply ingrained habits of structuring information, as you very correctly say.

If you think the phonetics of English are complicated, you should try German Or one of the many tonal languages, such as Thai, or Chinese…

Well, one must remember that Newtonian gravity is only a very simplified approximation that disregards all non-linearities, so it is perhaps not so surprising that some of its concepts turn out to be less general than we take them to be.

What Genady means is that the universe is a curved spacetime, as described by General Relativity. The thing now is that some concepts we are used to from old Newtonian physics do not straightforwardly translate to GR - and “mass” is one of them. The question of “what is the mass associated with a given region of spacetime” has no simple answer; there are in fact several different notions of mass that apply to different sets of circumstances, so it really depends. The underlying reasons for this is that the gravitational field in GR is self-coupling and thus itself a source of gravity (unlike in Newtonian gravity); but this type of energy cannot be localised, and is frame-dependent, so it is difficult to account for in an observer-independent way.

This is just combining two sources of gravity to obtain a new spacetime geometry - gravity nonetheless remains attractive in nature, as it always does for ordinary sources. But for the Alcubierre drive you need actual anti-gravity, which is a completely different thing - it can be shown in a general way that the Alcubierre metric in its original form requires exotic matter; unfortunately you cannot “cheat your way there” just by cleverly arranging ordinary sources. Also, even if you could construct an Alcubierre bubble, I think it would be completely unusable as a propulsion method, since it has some pretty nasty side effects and problems. Just a word of warning - these diagrams depicting spacetime curvature are just visual aids to understanding the basic concept, they are not accurate depictions of actual geometry. Spacetime geometry is intrinsic to the manifold, so there is no actual “direction” to curvature, since spacetime is not embedded into any kind of higher-dimensional space. The crucial concept to understand here is the distinction extrinsic vs intrinsic geometry; this is very important if you want to understand GR.

I disagree. While I know that Icelandic distinguishes these sounds in phonology and orthography, English effectively doesn’t, so there’s no point in this at all. Furthermore the vast majority of people here on this forum presumably will use English keyboards, so these letters are not straightforwardly accessible to them, making this not at all efficient to most of us. There are good reasons why English orthography is largely standardised, so deviating from this is unwise, really comes across as silly, and makes it hard for some to read your text. Trust me, this isn’t making a good impression. My advice to you, if you wish to engage in a proper discussion of your ideas and be taken seriously while doing so, is to stick to standard English orthography. You don’t have to agree with it, you just need to use it.

I disagree, I don’t think you have a bad reputation here. You’re still young, so there are a lot of things about modern physics which you haven’t encountered and learned about yet. We all understand this, so there’s no problem. The most important thing is to keep learning, and keep your mind open - don’t allow yourself to fall into the “I’ve figured this all out” trap. Trust me, the universe is far richer than you can even imagine right now. Keep learning